Sound Waves

SOUND WAVES IN SOLIDS

Sound waves can travel in solids just like they can travel in fluids. The speed of longitudinal sound waves in a solid rod can be shown to be

$v=\sqrt{Y/\rho }$

Where Y is the Young's modulus of the solid and $\rho$ its density. For extended solids, the speed is a more complicated function of bulk modulus and shear modulus. Table gives the speed of sound in some common materials.

Effect of Pressure, Temperature and Humidity on the speed of Sound in Air

We have stated that for an ideal gas, the pressure, volume and temperature of a given mass satisfy

$\frac{PV}{T}=cons\tan t$

As the density of a given mass is inversely proportional to its volume, the above equation may also be written as

$\frac{P}{\rho }=cT$

Where c is a constant. The speed of sound is

$v=\sqrt{\frac{\gamma P}{\rho }}=\sqrt{\gamma cT}$ …. (xvii)

Thus, if pressure is changed but the temperature is kept constant, the density varies proportionally and P/$\rho$ remains constant. The speed of sound is not affected by the change in pressure provided the temperature is kept constant.

If the temperature of air is changed then the speed of sound is also changed.

From equation (xvii),

$v\alpha \sqrt{T}$

At STP, the temperature is 0⁰ C or 273 K. If the speed of sound at 0⁰ C is v₀, its value at the temperature T (in Kelvin) will satisfy

$\frac{v}{v_0}=\sqrt{\frac{T}{273}}=\sqrt{\frac{273+t}{273}}$,

Where t is the temperature in ⁰C. This may be approximated as

$\frac{v}{v_0}={\left(1+\frac{t}{273}\right)}^{1/2}\cong 1+\frac{t}{546}$

or, $v=v_0=\left(1+\frac{t}{546}\right)$.

The density of water vapour is less than dry air at the same pressure. Thus, the density of moist air is less than that of dry air. As a result, the speed of sound increases with increasing humidity.

INTENSITY OF SOUND WAVES

As a wave travels in a medium, energy is tranorted from one part of the ace to another part. The intensity of a sound wave is defined as the average energy crossing a unit cross-sectional area perpendicular to the direction of propagation of the wave in unit time. It may also be stated as the average power transmitted across a unit cross-sectional area perpendicular to the direction of propagation.

The loudness of sound that we feel is mainly related to the intensity of sound. It also depends on the frequency to some extent.

Consider again a sound wave travelling along the x-direction. Let the equations for the dilacement of the particles and the excess pressure developed by the wave be given by

\begin{matrix}  {} \\  and \\\end{matrix}\,\,\,\,\,\,\,\,\,\left. \begin{align}  s\,=\,{{s}_{0}}\,\sin \,\omega \left( t\,-\,x/v \right) \\  p\,=\,{{p}_{0}}\,\cos \omega \,\left( t\,-\,x/v \right) \\ \end{align} \right| ….(xviii)

Where$P_0=\frac{B\omega s_0}{v}$.

Consider a cross-section of area A perpendicular to the x-direction. The medium to the left to it exerts a force pA on the medium to the right along the X-axis. The points of application of this force move longitudinally, that is along the force, with a speed $\frac{\partial s}{\partial t}$. Thus, the power W, transmitted by the wave from left to right across the cross-section considered, is $W=\left(pA\right)\frac{\partial s}{\partial t}$.

By (xviii),

W = Ap₀ cos$\omega$(t – x/v)$\omega$s₀ cos$\omega$(t – x/v)

$=\frac{A{\omega }^2{s}_0^{2}B}{v}{\cos }^2\omega \left(t-x/v\right)$.

The average of cos² $\omega$(t – x/v) over a complete cycle or over a long time is ½. The intensity I, which is equal to the average power transmitted across unit cross-sectional area is thus,

$I=\frac{1}{2}\frac{{\omega }^2{s}_0^{2}B}{v}=\frac{2{\pi }^{2}B}{v}{s}_0^2{v}^2$

$I=\frac{P_0^{2}V}{2B}$

As B = $\rho$v², the intensity can also be written as

$I=\frac{v}{2\rho v^2}{P}_0^2=\frac{P_0^2}{2\rho v}$

We see that the intensity is proportional to the square of the pressure amplitude P₀.

Loudness

Human ear is sensitive for extremely large range of intensity. So a logarithmic rather than an arithmetic scale is convenient. Accordingly, intensity level $\beta$ of a sound wave is defined by the equation

$\beta =10\log \left(\frac{I}{I_0}\right)$ decibel

Where I₀ = 10^–12 W/m² is the reference or threshold intensity level to which any intensity I is compared.

Illustration 1: Calculate the velocity of sound in air at N.T.P. The density of air at N.T.P. is 1.29 gm/ $\ell$. Assume air to be diatomic with $\gamma =\mathrm{1.4.}$Hence calculate the velocity of sound in air at 27⁰C.

Solution: Velocity of sound in air

= $\sqrt{\frac{\gamma P}{\rho }}$ = $\sqrt{\frac{1.4\times 1.013\times {10}^{5}N/m^2}{1.29kg/m^3}}$= 331.6m/s.

$\Rightarrow$ $\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}$

Using $\frac{P}{\rho }=\frac{R}{M}T$ $\Rightarrow$ v₂ = v₁ $\sqrt{\frac{T_2}{T_1}}=331.6\sqrt{\frac{273+27}{273}}$

We can see that the velocity of sound is proportional to the square root of absolute temperature.